# Is this a field in linear algebra?

Let $\alpha$ be a root of $x^4+4kx+1=0$ where $k$ is an integer. Is $\Bbb Q[\alpha]=\{a_0+a_1\alpha+a_2\alpha^2+a_3\alpha^3; a_i\in\Bbb Q\}$ a field?

I find it is quite hard to see $\Bbb Q[\alpha]$ is closed under the division operator. When I write $1/(a_0+a_1\alpha+a_2\alpha^2+a_3\alpha^3)=b_0+b_1\alpha+b_2\alpha^2+b_3\alpha^3$, I find it is not easy to find $b_i$.

• see the questions I asked in comments there math.stackexchange.com/questions/1718969/… to see why $\mathbb{Q}(c_1,\ldots,c_n)$ is a finite dimensional $\mathbb{Q}$ vector space when the $c_i$ are algebraic – reuns Apr 3 '16 at 3:46
• If you know a little bit of field theory and polynomial theory in abstract algebra, you can look at the answer below. If not, I think there is no easy elementary approach available. – Vim Apr 3 '16 at 3:53
• @Vim : the one I linked to. if $b \in L$ where $L = \mathbb{Q}[c_1,\ldots, c_n]$ where the $c_i$ are the roots of some irreducible polynomial of $\mathbb{Q}[x]$, then $b^k \in L$ for every $k \in \mathbb{Z}$. hence that vector space $L$ is a field – reuns Apr 3 '16 at 3:57
• @user1952009 still requires abstract algebra. I assume OP has only the knowledge of LA but not AA. – Vim Apr 3 '16 at 4:04
• @Vim : I don't agree, you don't need any AA knowledge, just by induction that every element $b$ in the vector space $L$ has a minimal polynomial, hence using the minimal polynomial for the $c_i$, the powers of $b$ are also in $L$ and also have a minimal polynomial. just induction, no need of complicated AA knowledge. and that proof is exactly the same as proving that the algebraic numbers are a field : it is a basic proof for constructing the rest of AA – reuns Apr 3 '16 at 4:11

You ought to look at it in the intended way: $\Bbb Q[\alpha]\cong \Bbb Q[x]/(x^4+4kx +1)$. If you show the polynomial is irreducible over $\Bbb Q$, then the quotient is a field. It's highly unlikely that anyone intends you to find inverses directly.
At the worst , you can brute force that question. Attempt to solve $(x^2+ax+b)(x^2+cx+d)=x^4+4kx+1$ for a factorization.